Symbol inside math operator

Question

I would like to define a math operator that looks a bit like this, which I mocked up in an image editor from the amsmath \square symbol:

Sometimes there will be a different letter than lambda inside the box, or an expression like \lambda', but the box should always be the same size, and always be square - it's meant to be an inline operator with a symbol drawn on it, not a box around part of a formula.

I don't mind if I have to manually position the formula inside it, as there will probably only be a few different letters that I'll use.

It doesn't have to be based on \square. In fact I think it would look nicer if the baseline was a bit lower, to match + for example.

Is there a reasonable way to achieve this?

Answer 1

Here's a solution using a \framebox of fixed size. I thought it would look better if the symbol in the square was smaller, in \scriptstyle. I think it looks reasonably good with any lowercase latin or greek letter.

\documentclass{article}
\usepackage{amsmath}
\newcommand{\squareop}[1]{%
    \setlength{\fboxsep}{0pt}%
    \setlength{\unitlength}{.7em}%
    \mathrel{%
        \raisebox{-1pt}{\framebox(1,1){\(\scriptstyle #1\)}}%
    }%
}
\begin{document}
\begin{tabular}{lll}
    \( x \squareop{a} y \)  & \( x \squareop{b} y \)    & \( x \squareop{c} y \)    \\
    \( x \squareop{d} y \)  & \( x \squareop{e} y \)    & \( x \squareop{f} y \)    \\
    \( x \squareop{g} y \)  & \( x \squareop{h} y \)    & \( x \squareop{i} y \)    \\
    \( x \squareop{j} y \)  & \( x \squareop{k} y \)    & \( x \squareop{l} y \)    \\
    \( x \squareop{m} y \)  & \( x \squareop{n} y \)    & \( x \squareop{o} y \)    \\
    \( x \squareop{p} y \)  & \( x \squareop{q} y \)    & \( x \squareop{r} y \)    \\
    \( x \squareop{s} y \)  & \( x \squareop{t} y \)    & \( x \squareop{u} y \)    \\
    \( x \squareop{v} y \)  & \( x \squareop{w} y \)    & \( x \squareop{z} y \)    \\
\end{tabular}
\qquad
\begin{tabular}{lll}
    \( x \squareop{\alpha} y \)     & \( x \squareop{\beta} y \)    & \( x \squareop{\gamma} y \)  \\
    \( x \squareop{\delta} y \)     & \( x \squareop{\epsilon} y \) & \( x \squareop{\zeta} y \)   \\
    \( x \squareop{\eta} y \)       & \( x \squareop{\theta} y \)   & \( x \squareop{\iota} y \)   \\
    \( x \squareop{\kappa} y \)     & \( x \squareop{\lambda} y \)  & \( x \squareop{\mu} y \)     \\
    \( x \squareop{\nu} y \)        & \( x \squareop{\xi} y \)      & \( x \squareop{\pi} y \)     \\
    \( x \squareop{\rho} y \)       & \( x \squareop{\sigma} y \)   & \( x \squareop{\tau} y \)    \\
    \( x \squareop{\upsilon} y \)   & \( x \squareop{\phi} y \)     & \( x \squareop{\varphi} y \) \\
    \( x \squareop{\chi} y \)       & \( x \squareop{\psi} y \)     & \( x \squareop{\omega} y \)  \\
\end{tabular}
\end{document}

Answer 2

You may also want the big version.

\documentclass{article}
\usepackage{amsmath,array,relsize}

\makeatletter
\DeclareRobustCommand{\boxop}[1]{\mathbin{\mathpalette\box@op{#1}}}
\DeclareRobustCommand{\bigboxop}[1]{\mathop{\mathpalette\bigbox@op{#1}}\slimits@}

\newcommand{\box@op}[2]{%
  \begingroup
  \sbox\z@{$\m@th#1\mkern15mu$}%
  \dimen@=\wd\z@
  \setlength{\fboxsep}{0pt}%
  \makebox[\dimen@]{%
    \framebox[0.9\dimen@]{%
      \vbox to 0.9\dimen@{%
        \vss
        \hbox{\raisebox{\depth}{$\box@op@style{#1}#2$}}%
        \vss
      }%
    }%
  }%
  \endgroup
}
\newcommand{\box@op@style}[1]{%
  \ifx#1\displaystyle\scriptstyle\else
  \ifx#1\textstyle\scriptstyle\else
  \scriptscriptstyle\fi\fi\m@th
}
\newcommand{\bigbox@op}[2]{%
  \begingroup
  \sbox\z@{$\m@th#1\sum$}%
  \dimen@=\wd\z@
  \vphantom{\sum}%
  \vcenter{%
    \setlength{\fboxsep}{0pt}%
    \hbox to \dimen@{%
      \hss
      \framebox[0.9\dimen@]{%
        \vbox to 0.9\dimen@{%
          \vss
          \hbox{\raisebox{\depth}{$\m@th#1\box@op@larger{#1}{#2}$}}%
          \vss
        }%
      }%
      \hss
    }%
  }%
  \endgroup
}
\newcommand{\box@op@larger}[2]{%
  \ifx#1\displaystyle
    \expandafter\@firstoftwo
  \else
    \expandafter\@secondoftwo
  \fi
  {\mathlarger{#2}}{#2}%
}
\makeatother

\begin{document}

$x\boxop{\lambda}y\boxop{\lambda\!'}z\boxop{\varphi}w\boxop{\beta}u$

$\scriptstyle x\boxop{\lambda}y\boxop{\lambda\!'}z$

$\displaystyle\sum_{k=1}^n\bigboxop{\lambda}_{k=1}^n x_k$
$\bigboxop{\lambda}_{k=1}^n x_k$
$\scriptstyle\bigboxop{\lambda}_{k=1}^n x_k$

$\displaystyle\sum_{k=1}^n\sum_{k=1}^n x_k$

\end{document}

Answer 3

Using youngtab that I not knew with the symbol inside math operator. It is right to write that the square has the same dimension.

\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amssymb}

\usepackage{youngtab}
\Ylinethick{0.5pt}
\begin{document}

$a \mathrel{\young(\lambda)} b$, $A_{\mathrel{\young(\mu)_c^d}}$, $X\sim\sum_{i=1}^nx_i\mathrel{\young(\diamond)}y_i$.

\end{document}